# Determining the transfer function of cruise control given characteristic equations

During my time revising for my control engineering module, i've come across this question

The automatic cruise speed control for a vehicle is described by the following equations

$\bigl( 1\bigl)$ $m \frac{dv_0(t)}{dt} + Fv_0(t)=u(t)$

$\bigl( 2\bigl)$ $u(t)=K[v_0(t)-v_r(t)]$

where m is the mass of the car, 1000$kg$, F is the frictional constant 50 $Nm^{-1}s$ and $u(t)$ is the throttle actuator force and K is the proportional gain control for the actuator. Also $V_0(t)$ is the actual speed while $V_r(t)$ is the desired speed.

There are two parts to this question, and im a bit confused on both of them.

part 1 says to draw a block diagram of the speed control system. In the control engineering module we generally draw the block diagrams as related to s, so i thought i'd get the laplace transform of the function (to get the transfer function) and draw it from the general model of a control system

This coincided with part 2, which wants me to show that the transfer function in terms of K is given by

$\frac{V_0(s)}{V_r(s)} = \frac{K}{1000s+50+K}$

I went about working it out and eventually ended up with this

$\frac{V_0(s)}{V_r(s)} = \frac{-K}{1000s+50-K}$

Can anyone help me figure out where i went wrong?

Also, i attempted making a block diagram in the t domain for this system, however since there are two equations with equal outputs (not summed) i was completely stumped on how i would display that as a model that is accurate, without making assumptions like equating K with the system gain of the other equation

• Our user flagged this question as a duplicate of this question at Math.SE. I'm a moderator at Math.SE. If you think we should do something about, @-ping me. We have a single zero score answer. Deleting our version may be a good idea. – Jyrki Lahtonen Jan 17 '18 at 14:12
• Why did you post this to two different sites without crosslinking? May be you didn't realize that we think that is a bit rude because it may result in people wanting to help you waste their time... – Jyrki Lahtonen Jan 17 '18 at 14:14
• @JackD'Aurizio I flagged the version in engineering. Their moderator responded with a request to migrate and flag for merging. I think I will just do that :-) – Jyrki Lahtonen Jan 18 '18 at 6:37

If you are using $(2)$ as your definition for $u(t)$ then your expression is correct. However in control it is often the convention to use negative feedback. In this case equation two would become

$$u(t) = K [v_r(t) - v_0(t)] \tag{2}$$

which is equivalent to negating your value of $K$. This would yield the same as the given result.

If it was me, and bearing in mind what the other answer said about negative feedback, I would do something like this:

Change the gain blocks from 1 to the correct parameters. vr is obviously your input/reference speed profile. I don't think there's any need to compute the transfer function to do a block diagram.

If you take the Laplace transform of the system you obtain:

$$msV_0(s)+FV_0(s)=U(s)$$ $$U(s)=K\left[V_0(s)-V_\text{r}(s) \right]$$

Plug the second equation into the first equation and solve for $V_0(s)/V_\text{r}(s)$:

$$\dfrac{V_0(s)}{V_\text{r}(s)}=\dfrac{-K}{ms+(F-K)},$$